Throughout this lesson we refer to two complex numbers $z_1=a+bi$ and $z_2=c+di$

We first need to establish the following result

$$2(ac+bd)=ac\mathbin{-}adi+bci+bd+ac+adi\mathbin{-}bci+bd=(a+bi)(c\mathbin{-}di)+(a\mathbin{-}bi)(c+di)=z_1\bar z_2+\bar z_1z_2$$

Now the square of the moduli of $z_1$ and $z_2$ are given by $|z_1|^2=a^2+b^2$ and $|z_2|^2=c^2+d^2$

And

$$z_1\pm z_2=(a\pm c)+(b\pm d)i$$

so

$$|z_1\pm z_2|^2=(a\pm c)^2+(b\pm d)^2= a^2+c^2+b^2+d^2\pm 2(ac+bd)=|z_1|^2+|z_2|^2\pm(z_1\bar z_2+\bar z_1 z_2)$$

using the result we established above

Thus in general

$$|z_1\pm z_2|^2\neq |z_1|^2+|z_2|^2$$

this result is of profound importance in Quantum Mechanics. For more information see Quantum Mechanics Lesson 7: Quantum Mysteries.

We ask the reader to show that 

$$|z_1z_2|=|z_1||z_2|$$

and

$$\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$$

This is straightforward algebra using the rectangular form and is trivial if you use the exponential form.

We now look at some properties of the conjugate which from the definition is given by

$$\bar z_1=a\mathbin{-}bi$$

and $$\bar z_2=c\mathbin{-}di$$

thus 

$$\bar z_1\pm\bar z_2=(a\pm c)\mathbin{-}(b\pm d)i$$

the sum or difference of the two complex numbers is given by

$$z_1\pm z_2=(a\pm c)+(b\pm d)i$$

and the conjugate of the sum or difference is thus

$$\overline{z_1\pm z_2}=(a\pm c)\mathbin{-}(b\pm d)i= \bar z_1\pm\bar z_2$$

in other words the conjugate of a sum or difference is the sum of difference of the conjugates.

we let the reader prove the corresponding statements for multiplication and division namely that the conjugate of the product or quotient of two complex numbers is the product or quotient of their conjugates:

$$\overline{z_1z_2}=\bar z_1\bar z_2$$

and

$$\overline{\left(\frac{z_1}{z_2}\right)}= \frac{\bar z_1}{\bar z_2}$$

Using the rectangular form the proof involves straightforward algebra and using the exponential form the proof is especially trivial.